One disadvantage to solving systems using substitution is that isolating a variable often involves dealing with messy fractions. There is another method for solving systems of equations: the addition/subtraction method.

In the addition/subtraction method, the two equations in the system are added or subtracted to create a new equation with only one variable. In order for the new equation to have only one variable, the other variable must cancel out. In other words, we must first perform operations on each equation until one term has an equal and opposite coefficient as the corresponding term in the other equation.

We can produce equal and opposite coefficients simply by multiplying each equation by an integer. Observe:

Example 1: Add and subtract to create a new equation with only one variable:


Here, we can multiply the second equation by -2:


Adding these two equations yields -2y = - 22.

Example 2: Add and subtract to create a new equation with only one variable:


Here, we can multiply the first equation by 3 and the second equation by 2:


Adding these two equations yields 26x = 78

We can add and subtract equations by the addition property of equality--since the two sides of one equation are equivalent, we can add one to one side of the second equation and the other to the other side.

Here are the steps to solving systems of equations using the addition/subtraction method:

1. Rearrange each equation so the variables are on one side (in the same order) and the constant is on the other side.
2. Multiply one or both equations by an integer so that one term has equal and opposite coefficients in the two equations.
3. Add the equations to produce a single equation with one variable.
4. Solve for the variable.
5. Substitute the variable back into one of the equations and solve for the other variable.
6. Check the solution--it should satisfy both equations.

Example 1: Solve the following system of equations:

1. Rearrange each equation:
   -3x + 2y = 7
   5x - 4y = - 12
2. Multiply the first equation by 2:
   -6x + 4y = 14
   5x - 4y = - 12
3. Add the equations:
   - x = 2
4. Solve for the variable:
   x = - 2
5. Plug x = - 2 into one of the equations and solve for y:
   -3(- 2) + 2y = 7
   6 + 2y = 7
   2y = 1
   y =
   Thus, the solution to the system of equations is (- 2,).
6. Check:
   2() - 3(- 2) = 7 ? Yes.
   5(- 2) = 4() - 12 ? Yes.

Example 2: Solve the following system of equations:

1. Rearrange each equation:
   3x + 4y = 25
   4x - 7y = - 16
2. Multiply the first equation by 4 and the second equation by -3:
   12x + 16y = 100
   -12x + 21y = 48
3. Add the equations:
   37y = 148
4. Solve for the variable:
   y = 4
5. Plug y = 4 into one of the equations and solve for x:
   3x + 4(4) = 25
   3x + 16 = 25
   3x = 9
   x = 3
   Thus, the solution to the system of equations is (3, 4).
6. Check:
   4(4) - 5 = 20 - 3(3) ? Yes.
   4(3) - 7(4) = - 16 ? Yes.

Example 3: Solve the following system of equations:

1. Rearrange each equation:
   2x - 5y = 15
   -4x + 10y = 20
2. Multiply the first equation by 2:
   4x - 10y = 30
   -4x + 10y = 20
3. Add the equations:
   0 = 50
   

Since 0≠50, this system of equations has no solutions. It is inconsistent (and independent). The equations describe two parallel lines.

Example 4: Solve the following system of equations:

1. Rearrange each equation:
   6x + 14y = 16
   -9x - 21y = - 24
2. Multiply the first equation by 3 and the second equation by 2:
   18x + 42y = 48
   -18x - 42y = - 48
3. Add the equations:
   0 = 0
   

Since 0 = 0 for any value of x, the system of equations has infinitely many solutions. Every ordered pair (x, y) which satisfies 6x + 14y = 16 (or -9x - 21y = - 24) is a solution to the system. The system is dependent (and consistent). The two equations describe the same line.